evaluate $\int_{0}^{1}\frac{\text{Li}_2(\frac{x^2-1}{4})}{1-x^2}dx$

Question

I came across this integral: $$\int_{0}^{1}\frac{\text{Li}_2(\frac{x^2-1}{4})}{1-x^2}dx=\frac{1}{2}\int_{-1}^{1}\frac{\text{Li}_2(\frac{x^2-1}{4})}{1-x^2}dx$$ One way to evaluate is to start with the subsitituion $x=2(v-0.5)$ to get: $$\int_{0}^{1}\frac{\text{Li}_2((v-1)v)}{v(1-v)}dv$$ We can use the series expansion of $\text{Li}_2((v-1)v)$ & beta function to get the A.A Markov series which equals $-\frac{4}{5}\zeta(3)$ which is: $$\sum_{n=1}^{\infty} \frac{2(-1)^{n}}{n^3\begin{pmatrix}2n\\ n\end{pmatrix}}$$ We can then prove that this sum equals $-\frac{4}{5}\zeta(3)$ by elementary means.

My question: is there anther approach to evaluate this integral besides using A.A Markov series? Like an identity of the dilogarithm function or integral manipulations?

Answer 1

Substitute $x=\sin y$ and we have \begin{align}\int_0^{\frac{\pi}{2}}\hbox{Li}_2\left(-\frac{\cos^2 y}{4}\right)\frac{dy}{\cos y}&=-\int_0^{\frac{\pi}{2}}\left(-\frac{\cos^2y}{4}\right)\int_0^1\frac{\log(t)}{1+\frac{t\cos^2 y}{4}}\frac{dtdy}{\cos y}\\ &=\int_0^1\log(t)\int_0^{\frac{\pi}{2}}\frac{\cos y}{4+t\cos^2 y}dydt\\&=\int_0^1\log(t)\int_0^{\frac{\pi}{2}}\frac{\cos y}{4+t\left(1-\sin ^2 y\right)}dydt\\&=\int_0^1\log(t)\tanh^{-1}\left(\sqrt{\frac{t}{t+4}}\right)\frac{dt}{\sqrt{t^2+4t}}\\&=4\int_0^1\log(t)\tanh^{-1}\left(\frac{t}{\sqrt{t^2+4}}\right)\frac{dt}{\sqrt{t^2+4}}\\&\overset {\text{IBP}}=-2\int_0^1\left(\tanh^{-1}\left(\frac{t}{\sqrt {t^2+4}}\right)\right)^2\frac{dt}{t}\\&=-2\int_0^1\frac{\hbox{arcsinh}^2\left(\frac{t}{2}\right)}{t}dt\\&=-2\int_0^{\frac{1}{2}}\frac{\hbox{arcsinh}^2(y)}{y}dy\end{align} The last integral is equal to series you have mentioned, however, for different approaches see here or here which is equal to $\frac{\zeta(3)}{10}$ and hence answer of original problem is $-\frac{\zeta(3)}{5}$. We used the following identities $$\hbox{Li}_2(z)=-z\int_0^1\frac{\log(t)}{1-zt}dt \\ \hbox{arcsinh}(x)=\hbox{arctanh}\left(\frac{x}{\sqrt {1+x^2}}\right)$$

Answer 2

Incomplete solution: \begin{align} J&=\int_{0}^{1}\frac{\text{Li}_2(\frac{x^2-1}{4})}{1-x^2}dx\\ J(a)&=\int_{0}^{1}\frac{\text{Li}_2(a(x^2-1))}{x^2-1}dx\\ J\left(0\right)&=0,J\left(\frac{1}{4}\right)=-J\\ J^\prime(a)&=\frac{1}{a}\int_0^1 \frac{\ln\big(1+a(1-x^2)\big)}{1-x^2}dx\\ J^{\prime}(a)&\overset{y=1-x^2}=\frac{1}{2a}\int_0^1 \frac{\ln(1+ay)}{y\sqrt{1-y}}dy\\ K(b)&=\int_0^1 \frac{\ln(1+by)}{y\sqrt{1-y}}dy\\ K(0)&=0\\ K^\prime(b)&=\int_0^1 \frac{1}{(1+by)\sqrt{1-y}}dy\\ &=-2\left[\frac{\text{arctanh}\left(\sqrt{\frac{b(1-y)}{1+b}}\right)}{\sqrt{b(1+b)}}\right]_{y=0}^{y=1}\\ &=\frac{2\text{arctanh}\left(\sqrt{\frac{b}{1+b}}\right)}{\sqrt{b(1+b)}}\\ K(b)&=K(b)-K(0)=\int_0^b K^\prime(t)dt=\int_0^b\frac{2\text{arctanh}\left(\sqrt{\frac{t}{1+t}}\right)}{\sqrt{t(1+t)}}dt\\ &\overset{u=\sqrt{\frac{t}{1+t}}}=\int_0^{\sqrt{\frac{b}{1+b}}}\frac{\text{4arctanh t}}{1-u^2}du\\ &=2\left(\text{arctanh}\left(\sqrt{\frac{b}{1+b}}\right)\right)^2\\ J^\prime(a)&=\frac{1}{2a}K(a)=\frac{\left(\text{arctanh}\left(\sqrt{\frac{a}{1+a}}\right)\right)^2}{a}\\ J&=-\left(J\left(\frac{1}{4}\right)-J(0)\right)=-\int_0^{\frac{1}{4}}\frac{\left(\text{arctanh}\left(\sqrt{\frac{a}{1+a}}\right)\right)^2}{a}da\\ &\overset{u=\sqrt{\frac{a}{1+a}}}=\int_0^{\frac{1}{\sqrt{5}}}\frac{2\text{artanh}^2 u}{u(u^2-1)}du\\ &=-\frac{1}{2}\int_0^{\frac{1}{\sqrt{5}}}\frac{\ln^2\left(\frac{1-u}{1+u}\right)}{u(1-u^2)}du\\ &\overset{z=\frac{1-u}{1+u}}=-\frac{1}{4}\int_{\frac{3-\sqrt{5}}{2}}^1\frac{(1+z)\ln^2 z}{(1-z)z}dz\\ &=-\frac{1}{4}\int_{\frac{3-\sqrt{5}}{2}}^1\frac{\ln^2 z}{z}dz-\frac{1}{2}\int_{\frac{3-\sqrt{5}}{2}}^1\frac{\ln^2 z}{1-z}dz\\ &\overset{\rho=\frac{3-\sqrt{5}}{2}}=\frac{1}{12}\ln^3\rho-\frac{1}{2}\left(2\text{Li}_3(1)-2\text{Li}_3(\rho)+2\text{Li}_2(\rho)\ln\rho+\ln(1-\rho)\ln^2\rho\right)\\ &=\boxed{\frac{1}{12}\ln^3\rho-\text{Li}_3(1)+\text{Li}_3(\rho)-\text{Li}_2(\rho)\ln\rho-\frac{1}{2}\ln(1-\rho)\ln^2\rho}\\ \end{align} Addendum: For the record:

\begin{align} J&=\int_{0}^{1}\dfrac{\text{Li}_2\left(\dfrac{x^2-1}{4}\right)}{1-x^2}dx\\ &\overset{x=\sin y}=\int_0^{\frac{\pi}{2}}\dfrac{\text{Li}_2\left(-\dfrac{1}{4}\cos^2 y\right)}{\cos y}dy\\ &=\int_0^{\frac{\pi}{2}}\left(\sum_{n=1}^\infty \frac{(-1)^n\cos^{2n-1}x}{2^{2n}n^2}\right)dy\\ &=\sum_{n=1}^\infty\left(\frac{(-1)^n}{2^{2n}n^2}\left(\int_0^{\frac{\pi}{2}}\cos^{2n-1}x dx\right)\right)\\ &=\sum_{n=1}^\infty\left(\frac{(-1)^n}{2^{2n}n^2}\times \frac{2^{2(n-1)}\big((n-1)!\big)^2}{(2n-1)!}\right)\\ &=\frac{1}{4}\sum_{n=1}^\infty\left(\frac{(-1)^n}{n^2}\times \frac{(n-1)!(n-1)!\times n\times 2n}{(2n-1)!\times n\times 2n}\right)\\ &=\frac{1}{2}\sum_{n=1}^\infty\frac{(-1)^n}{n^2}\times \frac{n!\times n!}{n(2n)!}\\ &=\frac{1}{2}\sum_{n=1}^\infty\frac{(-1)^n}{n^3\binom{2n}{n}} \end{align}